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Dirac Equation and commutation relations
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[QUOTE="stevendaryl, post: 5087247, member: 372855"] Well, the idea was that he wanted an equation of the form: [itex]H \psi = E \psi[/itex] and it had to be consistent with Special Relativity, which means that [itex]E^2 = p^2 c^2 + m^2 c^4[/itex] That implies that [itex]H = \sqrt{p^2 c^2 + m^2 c^4}[/itex]. But it's really difficult to work with square-roots. Besides that, in SR, energy and momentum are on the same footing, so Dirac reasoned that if the right side is linear in [itex]E[/itex], then the left side should be linear in [itex]p[/itex]. So he just guessed that it had the form: [itex]H = c \vec{p} \cdot \vec{\alpha} + \beta m c^2[/itex] for some constant vector [itex]\vec{\alpha}[/itex] and some constant [itex]\beta[/itex] Since [itex]H^2 \psi = E^2 \psi = (p^2 c^2 + m^2 c^4) \psi[/itex], the constants [itex]\vec{\alpha}[/itex] and [itex]\beta[/itex] had to satisfy certain conditions: [itex](c \vec{p} \cdot \vec{\alpha} + \beta m c^2)^2 = c^2 \sum_{i j} (p_i p_j) (\alpha_i \alpha_j) + m c^3 \sum_i p_i (\alpha_i \beta + \beta \alpha_i) + m^2 c^4 \beta^2[/itex] In order for this to equal [itex]c^2 p^2 + m^2 c^4[/itex], it must be that: [LIST=1] [*][itex]\alpha_i \alpha_j + \alpha_j \alpha_i = 0[/itex] if [itex]i \neq j[/itex] [*][itex]\alpha_i \alpha_i = 1[/itex] [*][itex]\alpha_i \beta + \beta \alpha_i = 0[/itex] [*][itex]\beta^2 = 1[/itex] [/LIST] The first two equations are obeyed by the Pauli spin matrices, which are 2x2. In order to accommodate the 3rd and 4th equations, he had to go to 4x4 matrices. [/QUOTE]
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Dirac Equation and commutation relations
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